I am given the following problem.

Let $Ω\subset \mathbb{C} $ and let $h \in C^1(Ω)$ be such that $\frac {\partial h} {\partial z}=0$. Show that $h(z)=\overline {f(z)}$ for some $f$ analytic in $Ω$.

The $\frac {\partial h} {\partial z}$ notation is mysterious to me. But, after looking at this, I have a guess as to what it means.

Write $h(x+iy)=u(x,y)+iv(x,y).$ Then $\frac {\partial h} {\partial z}=\frac {[u_x+iv_x]-i[u_y+iv_y]} 2=\frac {u_x+v_y+iv_x-iu_y} 2$. Is this interpretation correct?

  • $\begingroup$ It's just the ordinary derivative $\frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x} - \frac{\partial}{\partial y}\right)$. $\endgroup$ – anomaly Jan 24 at 23:16
  • $\begingroup$ @anomaly you can write that as an answer. It doesn't need to contain anything more $\endgroup$ – punctured dusk Apr 10 at 18:01

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